To determine the number of key comparisons that insertion sort makes to sort a list of 20 items when the list is given in reverse order, we need to analyze how insertion sort operates in this worst-case scenario.

Insertion Sort Overview

Insertion sort works by building a sorted portion of the list one element at a time. For each element, it compares it with the elements in the sorted portion and moves it to its correct position. When the list is in reverse order, every new element will have to be compared with all the elements in the sorted portion, resulting in the worst-case number of comparisons.

Calculating Key Comparisons

For a list of size $n = 20$, the insertion sort algorithm performs the following comparisons:

• For the 2nd element, 1 comparison is made.
• For the 3rd element, 2 comparisons are made.
• For the 4th element, 3 comparisons are made.
• ...
• For the last (20th) element, 19 comparisons are made.

Thus, the total number of comparisons $C(n)$ made by insertion sort for a list of $n$ elements in reverse order is:

$C(n) = 1 + 2 + 3 + \ldots + (n - 1)$

This is the sum of the first $n-1$ natural numbers, which can be calculated using the formula:

$C(n) = \frac{(n-1) \cdot n}{2}$

Applying the Formula

For $n = 20$:

$C(20) = \frac{(20-1) \cdot 20}{2} = \frac{19 \cdot 20}{2} = \frac{380}{2} = 190$

Conclusion

Insertion sort makes 190 key comparisons to sort a list of 20 items when the list is given in reverse order.

Would you like more details or have any other questions? Here are 5 related questions to consider:

1. What is the time complexity of insertion sort in the average case?
2. How does insertion sort compare to other sorting algorithms like bubble sort and selection sort?
3. How can we optimize insertion sort for partially sorted lists?
4. What are the advantages of using insertion sort for small datasets?
5. How many swaps does insertion sort make in the worst case?

Tip: Insertion sort is particularly efficient for small or nearly sorted datasets, making it useful in specific scenarios despite its $O(n^2)$ time complexity in the worst case.